Is PReLU superfluous with respect to ReLU?
Lets assume we have 3 Dense layers, where the activations are $x^0 \rightarrow x^1 \rightarrow x^2$, such that $x^2 = \psi PReLU(x^1) + \gamma$ and $x^1 = PReLU(Ax^0 + b)$   

Now lets see what it would take to conform the PReLU into a ReLU    

$\begin{align*} 
PReLU(x^1) &amp;= ReLU(x^1) + ReLU(p \odot x^1)\\
&amp;= ReLU(Ax^0+b) + ReLU(p\odot(Ax^0+b))\\
&amp;= ReLU(Ax^0+b) + ReLU((eye(p)A + eye(p)b)x^0)\\ 
&amp;= ReLU(Ax^0+b) + ReLU(Qx^0+c) \quad s.t. \quad Q = eye(p)A, \ \ c = eye(p)b\\
&amp;= [I, I]^T[ReLU(Ax^0+b), ReLU(Qx^0+c)]\\
\implies x^2 &amp;= [\psi, \psi][ReLU(Ax^0+b), ReLU(Qx^0+c)]\\
&amp;= V*ReLU(Sx^0 + d) \quad V=[\psi, \psi], \ \ S=[A, Q] \ \ d=[b, c]
\end{align*}$  

So as you said it is possible to break the form of the intermiediary $PReLU$ into a pure $ReLU$ while keeping it as a linear model, but if you take a second look at the parameters of the model, the size increase drastically. The hidden units of S doubled meaning to keep $x^2$ the same size $V$ also doubles in size. So this means if you dont want to use the $PReLU$ you are learning double the parameters to achieve the same capability (granted it allows you to learn a wider span of functions as well), and if you enforce the constraints on $V,S$ set by the $PReLU$ the number of paramaters is the same but you are still using more memory and more operations!  

I hope this example convinces you of the difference